A Level Further Mathematics 9231 Paper 23, Oct/Nov 2025

Evaluates determinant.

$k^2-7$

$k=\pm \sqrt{7}$

Applies chain rule: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}$

$=\frac{1+\sinh t}{\cosh t}$

Finds $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\cosh t(\cosh t)-(1+\sinh t)\sinh t}{{\cosh }^{2}t}$

$=\frac{1-\sinh t}{{\cosh }^{2}t}$

Applies chain rule: $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}$

$=\frac{1-\sinh t}{{\cosh }^{3}t}$

Finds Maclaurin's series with $y(0)$, $y^{\mathrm{\prime }}(0)$ and $y^{\mathrm{\prime }\mathrm{\prime }}(0)$ evaluated: $y=y(0)+x{y}^{\mathrm{\prime }}(0)+\frac{x^2}{2!}{y}^{\mathrm{\prime }\mathrm{\prime }}(0)$

$=1+x+\frac{1}{2}{x}^2$

Uses the product rule to differentiate $\frac{\mathrm{d}}{\mathrm{d}x}(x(1+x^2{)}^n)$

$=2n{x}^2(1+x^2{)}^{n-1}+(1+x^2{)}^n$

Uses $x^2=x^2+1-1$: $=2n(1+x^2-1)(1+x^2{)}^{n-1}+(1+x^2{)}^n$

Integrates both sides using the limits given: ${[x(1+x^2{)}^n]}_0^1=2n{I}_n-2n{I}_{n-1}+I_n$

Substitutes limits and rearranges: $2^n=(2n+1)I_n-2n{I}_{n-1}⟹(2n+1)I_n=2^n+2n{I}_{n-1}$

$I_{-1}={\left[{\tan }^{-1}x\right]}_0^1=\frac{1}{4}\pi$

Applies reduction formula with $n=-1$

$-I_{-1}=\frac{1}{2}-2I_{-2}$

$-\frac{1}{4}\pi =\frac{1}{2}-2I_{-2}⟹I_{-2}=\frac{1}{4}+\frac{1}{8}\pi$

Auxiliary equation: $5m^2+2m+1=0⟹m=-\frac{1}{5}\pm \frac{2}{5}\mathrm{i}$

Complementary function: $y={\mathrm{e}}^{-\frac{1}{5}x}(A\cos \frac{2}{5}x+B\sin \frac{2}{5}x)$

Particular integral and its derivatives: $y=p{x}^2+qx+r⟹y^{\mathrm{\prime }}=2px+q⟹y^{\mathrm{\prime }\mathrm{\prime }}=2p$

Substitutes and equates coefficients: $p=14p+q=510p+2q+r=3$

$q=1r=-9$

General solution: $y={\mathrm{e}}^{-\frac{1}{5}x}(A\cos \frac{2}{5}x+B\sin \frac{2}{5}x)+x^2+x-9$

Differentiates using the product rule: $y^{\mathrm{\prime }}={\mathrm{e}}^{-\frac{1}{5}x}(-\frac{2}{5}A\sin \frac{2}{5}x+\frac{2}{5}B\cos \frac{2}{5}x)-\frac{1}{5}{\mathrm{e}}^{-\frac{1}{5}x}(A\cos \frac{2}{5}x+B\sin \frac{2}{5}x)+2x+1$

Uses initial conditions: $A-9=0\frac{2}{5}B-\frac{1}{5}A+1=0$

$⟹A=9,B=2$

$y={\mathrm{e}}^{-\frac{1}{5}x}(9\cos \frac{2}{5}x+2\sin \frac{2}{5}x)+x^2+x-9$

Forms the sum of the areas of the rectangles with correct width and at least three complete terms including first and last.

$\left(\frac{1}{n}\right)(\frac{1}{3}{\left(\frac{1}{n}\right)}^3+\left(\frac{1}{n}\right))+\left(\frac{1}{n}\right)(\frac{1}{3}{\left(\frac{2}{n}\right)}^3+\left(\frac{2}{n}\right))+\cdots +\left(\frac{1}{n}\right)(\frac{1}{3}{\left(\frac{n}{n}\right)}^3+\left(\frac{n}{n}\right))$

Applies formulae from MF19: $\frac{1}{3n^4}{\sum }_{r=1}^n{r}^3+\frac{1}{n^2}{\sum }_{r=1}^{n}r$

$=\frac{n^2(n+1{)}^2}{12n^4}+\frac{n(n+1)}{2n^2}$

$=\frac{1}{12}{(1+\frac{1}{n})}^2+\frac{1}{2}(1+\frac{1}{n})=\frac{1}{12}(1+\frac{1}{n})(7+\frac{1}{n})$

Forms the sum of the areas of appropriate rectangles.

$\left(\frac{1}{n}\right)(\frac{1}{3}{\left(\frac{1}{n}\right)}^3+\left(\frac{1}{n}\right))+\left(\frac{1}{n}\right)(\frac{1}{3}{\left(\frac{2}{n}\right)}^3+\left(\frac{2}{n}\right))+\cdots +\left(\frac{1}{n}\right)(\frac{1}{3}{\left(\frac{n-1}{n}\right)}^3+\left(\frac{n-1}{n}\right))$

Applies formulae from MF19: $\frac{1}{3n^4}{\sum }_{r=1}^{n-1}{r}^3+\frac{1}{n^2}{\sum }_{r=1}^{n-1}r=\frac{(n-1{)}^2{n}^2}{12n^4}+\frac{(n-1)n}{2n^2}$

$=\frac{1}{12}{(1-\frac{1}{n})}^2+\frac{1}{2}(1-\frac{1}{n})=\frac{1}{12}(1-\frac{1}{n})(7-\frac{1}{n})$

Simplifies $U_n-L_n$ to $\frac{c}{n}$: $\frac{1}{12}\left(\frac{16}{n}\right)=\frac{4}{3n}$

$\frac{4}{3n}\to 0\text{ as }n\to \mathrm{\infty }$

$a,2a,-3a$

$${\mathbf{P}}^2=\left(\begin{array}{ccc}{a}^2& 3a& -2a-1\\ 0& 4a^2& a\\ 0& 0& 9a^2\end{array}\right)$$

States that $\mathbf{P}$ satisfies its characteristic equation: ${\mathbf{P}}^3-7a^2\mathbf{P}+6a^3\mathbf{I}=\mathbf{0}$

Multiplies through by ${\mathbf{P}}^{-1}$: $6a^3{\mathbf{P}}^{-1}=7a^2\mathbf{I}-{\mathbf{P}}^2$

$${\mathbf{P}}^{-1}=\frac{1}{6a^3}\left(\begin{array}{ccc}6a^2& -3a& 2a+1\\ 0& 3a^2& -a\\ 0& 0& -2a^2\end{array}\right)$$

$$\mathbf{D}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)$$

Applies $\mathbf{A}=\mathbf{P}\mathbf{D}{\mathbf{P}}^{-1}$

Multiplies two adjacent matrices. For M1, $\mathbf{P}$ must be correct.

$$=\frac{1}{6a^3}\left(\begin{array}{ccc}a& 2& 3\\ 0& 4a& -3\\ 0& 0& -9a\end{array}\right)\left(\begin{array}{ccc}6a^2& -3a& 2a+1\\ 0& 3a^2& -a\\ 0& 0& -2a^2\end{array}\right)$$

$$\left(\begin{array}{ccc}1& \frac{1}{2a}& -\frac{4a+1}{6a^2}\\ 0& 2& \frac{1}{3a}\\ 0& 0& 3\end{array}\right)$$

Applies ${\tanh }^{-1}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$ or implicit differentiation $\tanh y=x⟹{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}y\frac{\mathrm{d}y}{\mathrm{d}x}=1$

$\frac{1}{2}\left(\frac{1-x}{1+x}\right)\frac{(1-x)(1)-(1+x)(-1)}{(1-x{)}^2}$ or $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{1-{\tanh }^{2}y}$

$=\frac{1}{1-x^2}$

Divides through by $x$: $\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{y}{x}=x{\tanh }^{-1}x$

Finds integrating factor: ${\mathrm{e}}^{-\int x^{-1}\mathrm{d}x}$

$={\mathrm{e}}^{-\ln x}=x^{-1}$

Correct form on LHS and RHS: $\frac{\mathrm{d}}{\mathrm{d}x}(x^{-1}y)={\tanh }^{-1}x$

Integrates correct function on RHS: $x^{-1}y=x{\tanh }^{-1}x-\int \frac{x}{1-x^2}\mathrm{d}x$

Correct integration of RHS

$x^{-1}y=x{\tanh }^{-1}x+\frac{1}{2}\ln (1-x^2)+C$

Substitutes initial conditions: $0=\frac{1}{2}{\tanh }^{-1}\frac{1}{2}+\frac{1}{2}\ln \frac{3}{4}+C⟹C=-\frac{1}{4}\ln 3-\frac{1}{2}\ln \frac{3}{4}$

$x^{-1}y=x{\tanh }^{-1}x+\frac{1}{2}\ln (1-x^2)-\frac{1}{4}\ln \frac{27}{16}$

$$\frac{z^{n+1}-z}{z-1}$$

Expresses $\frac{z^{n+1}-z}{z-1}$ in terms of $\theta$

Multiplies numerator and denominator by complex conjugate of $z-1$.

Takes imaginary part.

Obtains correct expression for imaginary part before using compound angle formula

Uses $\sin (A-B)=\sin A\cos B-\cos A\sin B$ in numerator

$=\frac{{\left(\frac{1}{2}\right)}^{n+2}\sin n\theta -{\left(\frac{1}{2}\right)}^{n+1}\sin (n+1)\theta +\frac{1}{2}\sin \theta }{\frac{5}{4}-\cos \theta }$

Differentiates the result from (b): ${\sum }_{m=1}^n{\left(\frac{1}{2}\right)}^{m}m\cos m\theta =\frac{\mathrm{d}}{\mathrm{d}\theta }(\dots )$

Applies quotient rule or product rule to differentiate the fraction

$=\frac{(\frac{5}{4}-\cos \theta )({\left(\frac{1}{2}\right)}^{n+2}n\cos n\theta -{\left(\frac{1}{2}\right)}^{n+1}(n+1)\cos (n+1)\theta +\frac{1}{2}\cos \theta )-({\left(\frac{1}{2}\right)}^{n+2}\sin n\theta -{\left(\frac{1}{2}\right)}^{n+1}\sin (n+1)\theta +\frac{1}{2}\sin \theta )\sin \theta }{{(\frac{5}{4}-\cos \theta )}^2}$

Uses $n{\left(\frac{1}{2}\right)}^n\to 0$ as $n\to \mathrm{\infty }$: $\frac{(\frac{5}{4}-\cos \theta )(\frac{1}{2}\cos \theta )-(\frac{1}{2}\sin \theta )\sin \theta }{{(\frac{5}{4}-\cos \theta )}^2}$

$=\frac{\frac{5}{8}\cos \theta -\frac{1}{2}}{{(\frac{5}{4}-\cos \theta )}^2}$